Problem: Divide the following complex numbers. $ \dfrac{1+21i}{1+4i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-4i}$ $ \dfrac{1+21i}{1+4i} = \dfrac{1+21i}{1+4i} \cdot \dfrac{{1-4i}}{{1-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(1+21i) \cdot (1-4i)} {(1+4i) \cdot (1-4i)} = \dfrac{(1+21i) \cdot (1-4i)} {1^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(1+21i) \cdot (1-4i)} {(1)^2 - (4i)^2} = $ $ \dfrac{(1+21i) \cdot (1-4i)} {1 + 16} = $ $ \dfrac{(1+21i) \cdot (1-4i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1+21i}) \cdot ({1-4i})} {17} = $ $ \dfrac{{1} \cdot {1} + {21} \cdot {1 i} + {1} \cdot {-4 i} + {21} \cdot {-4 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{1 + 21i - 4i - 84 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{1 + 21i - 4i + 84} {17} = \dfrac{85 + 17i} {17} = 5+i $